HYDRAULICS 355 



with one end contraction, n = 1 ; and for a weir with no end con- 

 tractions, n = 0. In the last case, the two preceding formulas 

 become, respectively, 



Q = 3.33bH$ 

 and Q = 3.33b[(H+h)%-h$] 



The velocity of approach can be determined by first finding 

 Q from the formula Q = 3.33 bH%, and dividing it by the area 

 of the cross-section of the channel; the quotient will be the 

 velocity of approach v and h will equal .01555 v 2 . 



EXAMPLE 1. A weir with end contractions is 5 ft. long and 

 the measured head is .872 ft. Calculate the discharge on the 

 assumption that the velocity of approach is negligible. 



SOLUTION. Substituting the given values in the proper 

 formula. Q = 3.33 X (5 -AX. 872) X. 872 *= 13.085 cu. ft. per 

 sec. 



The preceding formulas are known as Francis's formulas and 

 are recommended for heads from 5 to 19 in. For lower heads, the 

 formula of Fteley and Stearns, which follows, i? recommended: 

 Q = 3.31fc(tf-f 1.5/i)*+. 007& 



For higher heads, Bazin's formula is recommended: 



The last two formulas are applicable only to weirs with no 

 end contractions. In these formulas, p is the distance from 

 the bottom of the channel to the crest; the other letters have 

 the same significance as before. 



Triangular Weir. The form 

 of weir shown in Fig. 2 may be 

 used for small flows where the 

 head lies between the limits of 

 .02 and 1 ft. For a right-angled 

 weir with sharp inner edges, 



EXAMPLE . Calculate the 

 discharge of a triangular weir whose effective head is 9 in. 

 SOLUTION. Substituting the given values in the formula, 

 Q = 2.54 X.75* = 1.24 cu. ft. per sec. 



